Solve below somewhat symmetric equations: x, y, z subject to x^2+y^2 -

Yaritza Robinson 2022-04-21 Answered
Solve below somewhat symmetric equations:
x, y, z subject to
x2+y2xy=3
(xz)2+(yz)2(yz)(xz)=4
(xz)2+y2y(xz)=1
x,y,zR+
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Answers (2)

Volsa280
Answered 2022-04-22 Author has 16 answers

Step 1
We can take a quadrilateral ABCD, for which:
ACBD={E}, AC=BD=z, ED=x, EC=y, AEB=60°, AB=2, BC=1 and CD=3
Thus, we obtain your conditions:
x2+y2xy=ED2+EC22EDECcos60=CD2=3,
(xz)2+(yz)2(yz)(xz)=(zx)2+(zy)2(zx)(zy)=
=EB2+EA22EBEAcos60=AB2=4
and(xz)2+y2y(xz)=(zx)2+y2+y(zx)=
=EB2+EC22EBECcos120=1.
Now, let BFCD be a parallelogram.
Thus,
AC=z=BD=FC
and
ACF=DEC=60°,
which gives that AFC is equilateral triangle, BA=2, BF=CD=3 adn BC=1
Now, let R be a rotation around C by 60 and R(B)=G thus, since BGC is an equilateral triangle, we obtain: BG=1, FG=R(AB)=2 and
FG2=22=(3)2+12=BF2+BC2,
which says
FBG=90°,
FBC=FBC+GBC=90°+60°=150°,
which gives
z2=AC2=FC2=FB2+BC22FBBCcos150=3+1+3=7,
whisch gives
z=7
and from here we get the answer:
(x,y,z)=(57,17,7).

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Felicity Carter
Answered 2022-04-23 Author has 16 answers
Step 1
We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.
[(xz)2+(yz)2(yz)(xz)]
4[(xz)2+y2y(xz)]=0
x=34y2+y2
y=143313
z=±3
It should be easy to figure out x from here.
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